Problem: The roots of (x2β3x+2)(x)(xβ4)=0\left(x^{2}-3 x+2\right)(x)(x-4)=0(x2β3x+2)(x)(xβ4)=0 are x=x=x=:
Answer Choices:
A. 444
B. 000 and 444
C. 111 and 222
D. 0,1,20,1,20,1,2 and 444
E. 1,21, 21,2 and 444 Solution:
Set each of the factors equal to zero: x=0,xβ4=0,x2β3x+2=(xβ2)β (xβ1)=0;β΄x=0,4,2,1.x=0, x-4=0, x^{2}-3 x+2=(x-2) \cdot(x-1)=0 ; \therefore x=0,4,2,1 .x=0,xβ4=0,x2β3x+2=(xβ2)β (xβ1)=0;β΄x=0,4,2,1.